3. The attempt at a solution
I am able to prove that y[n] = x[n]cos(0.2*pi*n) is linear by saying
let x[n] = ax1[n] + bx2[n]
then y[n] = (ax1[n] + bx2[n])cos(0.2*pi*n) = ax1[n]cos(0.2*pi*n) + bx2[n]cos(0.2*pi*n)
so y[n] = ay1[n] + by2[n] proving that it is linear

I am not able to prove, at this point, that y[n] is either time-invariant or not.
All I have is this:
let g[n] = x[n - d]
then y[n - d] = g[n]cos(0.2*pi*n) = x[n - d]cos(0.2*pi*n), so it is time-invariant?
This doesn't seem right though for whatever reason.

The system is time invariant if it gives the same output regardless of whether a delay is made to the input signal, or to the output signal.
So, given
y[n] = x[n]cos(0.2*pi*n)
If the output signal is delayed,
y[n-d] = x[n-d]cos(0.2*pi*(n-d))
However, if the delay was made to the input signal, before inputing it into the system ... Can you take it from here?